Problem 18 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int_{0}^{\pi / 4} x \cos 2 x d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi - 2}{8} \).

Step 1: Identify Parts for Integration by Parts

In the integral \( \int x \cos(2x) \ dx \), we use the formula for integration by parts: \( \int u \ dv = uv - \int v \ du \). Identify \( u \) as \( x \) and \( dv \) as \( \cos(2x) \ dx \). This means \( du = dx \) and to find \( v \), integrate \( dv \).

Step 2: Find the Integral of dv

To find \( v \), integrate \( dv = \cos(2x) \ dx \). The integral is \( v = \frac{1}{2} \sin(2x) \). This results from using the substitution method or recognizing the derivative of \( \sin(2x) \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula. This gives: \( \int x \cos(2x) \ dx = x \left( \frac{1}{2} \sin(2x) \right) - \int \left( \frac{1}{2} \sin(2x) \right) \ dx \).

Step 4: Simplify and Integrate

Simplify the expression: \( \frac{x}{2} \sin(2x) - \frac{1}{2} \int \sin(2x) \ dx \). The next task is to integrate \( \sin(2x) \).

Step 5: Integrate \( \sin(2x) \)

To integrate \( \sin(2x) \), we recognize that the integral is \( -\frac{1}{2} \cos(2x) \). This is because the derivative of \( \cos(2x) \) is \(-2 \sin(2x) \).

Step 6: Finalize the Indefinite Integral

Combine the results to get: \( \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \). This combines both parts of the integration by parts formula.

Step 7: Evaluate Definite Integral

Evaluate the expression from \( 0 \) to \( \frac{\pi}{4} \). Substituting, we find: \( \left[ \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \right]_{0}^{\pi/4} \).

Step 8: Plug in Upper Bound

Calculate the expression at \( x = \frac{\pi}{4} \): \( \frac{\frac{\pi}{4}}{2} \sin\left(\frac{\pi}{2}\right) + \frac{1}{4} \cos\left(\frac{\pi}{2}\right) = \frac{\pi}{8} \times 1 + 0 \).

Step 9: Plug in Lower Bound

Calculate the expression at \( x = 0 \): \( \frac{0}{2} \sin(0) + \frac{1}{4} \cos(0) = 0 + \frac{1}{4} \).

Step 10: Subtract Lower Bound Result

Subtract the lower bound result from the upper bound result: \( \frac{\pi}{8} - \frac{1}{4} = \frac{\pi}{8} - \frac{2}{8} = \frac{\pi - 2}{8} \).

Key Concepts

Definite Integrals

Definite integrals involve finding the area under a curve within a specific range, from one point to another on the x-axis. Here, we look at how to calculate \[\int_{a}^{b} f(x) \, dx\] using the integration by parts technique. In the given problem, the integrand is a combination of algebraic and trigonometric functions.
Integration by parts is similar to the product rule for derivatives, and is used when you have the product of two different types of functions. For definite integrals, once you have the indefinite integral, you evaluate it at the specified bounds \[a\] and \[b\].

The evaluation step is crucial because it transforms the indefinite integral into a definite value that represents an actual area or total accumulated quantity. This can help visualize how changes in the function across a given interval impact the total integral value.

Trigonometric Integrals

Trigonometric integrals involve functions like \( \sin(x)\) and \( \cos(x)\). These integrals come up frequently in calculus, especially when combined with other functions, such as polynomials, which require techniques like integration by parts.
In the example, one part of the integral involves the trigonometric \( \cos(2x) \), which later converts to \( \sin(2x) \) when integrated. Recognizing and manipulating these trigonometric forms can help simplify the integration process, especially when used with identities and substitution methods.

When dealing with trigonometric integrals, remember that transformations often involve using basic trigonometric identities or finding derivatives and antiderivatives of trigonometric functions. This knowledge helps in simplifying the integral or changing it into a form that is easier to handle.

Calculus Techniques

Integration by parts is a critical technique in calculus, useful for integrating the product of two functions. This method applies the formula:

• \[ \int u \, dv = uv - \int v \, du \]

to separate the functions and derive their antiderivative.
We start by identifying which portion to set as \( u \) (the simpler function for differentiation) and \( dv \) (what we can manage to integrate). Choosing wisely here can streamline the process considerably.

In our original exercise, we chose \( u = x \) and \( dv = \cos(2x) \, dx \). Post integration, remember to always apply the bounds, if you're dealing with definite integrals. This wraps up the calculation and gives you the numeric value for an area or total change described by the function between given limits.